Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Solution {
    static MyScanner sc;
    private static PrintWriter out;
    static long M2 = 1_000_000_000L + 7;

    public static void main(String[] s) throws Exception {
        StringBuilder stringBuilder = new StringBuilder();
//        stringBuilder.append("100000 1000000000 10001 100000 ");
//        for (int i = 1; i < 100000; i++) {
//            stringBuilder.append(" 10000 100000 ");
//        }
        if (stringBuilder.length() == 0) {
            sc = new MyScanner(System.in);
        } else {
            sc = new MyScanner(new BufferedReader(new StringReader(stringBuilder.toString())));
        }

        out = new PrintWriter(new OutputStreamWriter(System.out));
        initData();
        solve();
        out.flush();
    }


    private static void initData() {

    }

    private static void solve() throws IOException {
        int n = sc.nextInt();
        int m = sc.nextInt();
        int[][] data = new int[n][];
        for (int i = 0; i < n; i++) {
            data[i] = sc.na(m);
        }
        ArrayList<String> tt = new ArrayList<>();
        if (n < m) {
            tt.addAll(k1(data));
            tt.addAll(k2(data));
        } else {
            tt.addAll(k2(data));
            tt.addAll(k1(data));
        }
        long s = Arrays.stream(data).flatMap(t -> Arrays.stream(t).boxed()).distinct().count();
        if (s != 1) {
            out.println(-1);
            return;
        }
        out.println(tt.size());
        for (String mm : tt) {
            out.println(mm);
        }
    }

    private static Collection<? extends String> k2(int[][] data) {
        ArrayList<String> res = new ArrayList<>();
        for (int s = 0; s < data[0].length; s++) {
            int min = 1000;
            for (int i = 0; i < data.length; i++) {
                min = Math.min(min, data[i][s]);
            }
            for (int l = 0; l < min; l++) {
                res.add("col " + (s + 1));
            }
            for (int i = 0; i < data.length; i++) {
                data[i][s] -= min;
            }

        }

        return res;
    }

    private static Collection<? extends String> k1(int[][] data) {
        ArrayList<String> res = new ArrayList<>();
        for (int s = 0; s < data.length; s++) {
            int min = 1000;
            for (int i = 0; i < data[0].length; i++) {
                min = Math.min(min, data[s][i]);
            }
            for (int l = 0; l < min; l++) {
                res.add("row " + (s + 1));
            }
            for (int i = 0; i < data[0].length; i++) {
                data[s][i] -= min;
            }

        }

        return res;
    }


    private static void solveT() throws IOException {
        int t = sc.nextInt();
        while (t-- > 0) {
            solve();
        }
    }

    private static long gcd(long l, long l1) {
        if (l > l1) return gcd(l1, l);
        if (l == 0) return l1;
        return gcd(l1 % l, l);
    }

    private static long pow(long a, long b, long m) {
        if (b == 0) return 1;
        if (b == 1) return a;
        long pp = pow(a, b / 2, m);
        pp *= pp;
        pp %= m;
        return (pp * (b % 2 == 0 ? 1 : a)) % m;
    }


    static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        MyScanner(BufferedReader br) {
            this.br = br;
        }

        public MyScanner(InputStream in) {
            this(new BufferedReader(new InputStreamReader(in)));
        }

        void findToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
        }

        String next() {
            findToken();
            return st.nextToken();
        }

        Integer[] nab(int n) {
            Integer[] k = new Integer[n];
            for (int i = 0; i < n; i++) {
                k[i] = sc.fi();
            }
            return k;
        }

        int[] na(int n) {
            int[] k = new int[n];
            for (int i = 0; i < n; i++) {
                k[i] = sc.fi();
            }
            return k;
        }

        long[] nl(int n) {
            long[] k = new long[n];
            for (int i = 0; i < n; i++) {
                k[i] = sc.nextLong();
            }
            return k;
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int fi() {
            String t = next();
            int cur = 0;
            boolean n = t.charAt(0) == '-';
            for (int a = n ? 1 : 0; a < t.length(); a++) {
                cur = cur * 10 + t.charAt(a) - '0';
            }
            return n ? -cur : cur;
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }


}